Finding the limit $\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}$

Question

I am trying to find this limit

$$\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}$$

My so far method is this

• $f(x)>0.$
• $f^{\prime}(x)=\frac{1}{3\sqrt[3]{(x+1)^2}}-\frac{1}{3\sqrt[3]{x^2}}<0.$
• For every $0<y<1$ the equation $f(x)=y$ has a unique solution(found in maple):

$$x=\frac{1}{3}\cdot \frac{\frac{1}{3}y^2(3y^2+\sqrt{-3y^4+12y})+\frac{1}{6}\frac{3y^2+\sqrt{-3y^4+12y}}{y}-y^4-2y}{y}$$

The previous facts implies that the limit is $0$.

I am wondering if there is any way easier than this to find the limite.

thanks in advance.

Answer 1

$$F=\lim_{x\to\infty^+}(\sqrt[n]{x+1}-\sqrt[n]x)$$

$$=\lim_{h\to0^+}\frac{\sqrt[n]{1+h}-1}{\sqrt[n]h}$$

Now for the numerator use $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})$ to find

$$F=\lim_{h\to0^+}\frac{1+h-1}{\sqrt[n]h}\cdot\frac1{\lim_{h\to0^+}[(\sqrt[n]{1+h})^{n-1}+(\sqrt[n]{1+h})^{n-2}+(\sqrt[n]{1+h})+1]}$$

Answer 2

use the binomial theorem to expand $$(x+1)^{1/3} = x^{1/3} + \frac 13 x^{-2/3}+\cdots $$ so $$ (x+1)^{1/3} - x^{1/3} = \frac 13 x^{-2/3}+\cdots \rightarrow 0\text{ as } x \to \infty.$$

Answer 3

Mean value theorem. If $f(x)=x^{1/3}$ then $f'(x)=\frac{1}{3}x^{-2/3}$.

Now, $f(x+1)-f(x)=1\cdot f'(c)$ for some $c\in[x,x+1]$. But $f'$ is decreasing and positive for $x>0$, so $$0<f(x+1)-f(x)\leq \frac{1}{3}x^{-2/3}.$$ By the squeeze theorem, $f(x+1)-f(x)\to 0$.

More generally, if $f$ is any function that is differentiable such that $\lim_{x\to\infty} f'(x)$ exists, then $$\lim_{x\to\infty} f(x+a)-f(x)=a\lim_{x\to\infty} f'(x).$$

Answer 4

hint:Let $x = \dfrac{1}{h} \to x \to \infty \iff h \to 0 \to L = \displaystyle \lim_{h \to 0} \dfrac{\sqrt[3]{1+h} - 1}{h}\cdot \displaystyle \lim_{h\to 0}\dfrac{h}{\sqrt[3]{h}}$

Answer 5

Yet another method: let $a = (x+1)^{1/3}$, $b = x^{1/3}$, so $$f(x) = (x+1)^{1/3} - x^{1/3} = a - b = \frac{(a-b)(a^2 + ab + b^2)}{a^2 + ab + b^2} = \frac{a^3 - b^3}{a^2 + ab + b^2}.$$ But $a^3 - b^3 = (x+1) - x = 1$, hence $$f(x) = \frac{1}{a^2 + ab + b^2}.$$ Now since $a \to \infty$ and $b \to \infty$ as $x \to \infty$, it is obvious that $$\lim_{x \to \infty} f(x) = 0.$$

Answer 6

Hint:-

$$0<\left(\sqrt[3]{x+1}-\sqrt[3]{x}\right)=\displaystyle\frac{1}{\left(\sqrt[3]{(x+1)^2}+\sqrt[3]{x(x+1)}+\sqrt[3]{x^2}\right)}<\displaystyle\frac{1}{3\sqrt[3]{x^2}}$$